6.3 Complex Eigenvalues and Eigenvectors

1. Given $z=1+i$ and $w=2-3i$, compute the following complex numbers.

a. $z+w$

$$z+w=(1+i)+(2-3i)=3-2i$$

b. $2z$

$$2z=2(1+i)=2+2i$$

c. $zw$

$$zw=(1+i)(2-3i)=2-3i+2i-3i^2=2-i+3=5-i$$

d. $|z|$

$$|z|=\sqrt{1^2+1^2}=\sqrt{2}$$

e. $z\bar{z}$

$$z\bar{z}=(1+i)(1-i)=1-i^2=1+1=2$$

f. $\frac{1}{w}$

$$\frac{1}{w}=\frac{1}{2-3i}\cdot \frac{2+3i}{2+3i}=\frac{2+3i}{4+9}=\frac{2+3i}{13}$$

2. Find the polar form of $z=\sqrt{3}+i$ and use it to compute $z^2$.

Polar Form

$$r=\sqrt{(\sqrt{3}{)}^2+1^2}=2$$ $$\theta =\arctan \left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$$ $$z=2\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)$$

Compute $z^2$

$$z^2={\left[2\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)\right]}^2=4\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)=2+2\sqrt{3}i$$

3. Find the eigenvalues and associated eigenvectors of the matrix $A=\left[\begin{array}{cc}2& -2\\ 3& 2\end{array}\right]$.

Eigenvalues

$$\det (A-\lambda I)=\left|\begin{array}{cc}2-\lambda & -2\\ 3& 2-\lambda \end{array}\right|=(2-\lambda {)}^2+6={\lambda }^2-4\lambda +10=0$$ $$\lambda =2\pm 2i$$

Eigenvectors

For $\lambda =2+2i$:

$$\left[\begin{array}{cc}-2i& -2\\ 3& -2i\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0⟹x=iy$$

Eigenvector: $\left[\begin{array}{c}1\\ i\end{array}\right]$

For $\lambda =2-2i$:

$$\left[\begin{array}{cc}2i& -2\\ 3& 2i\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0⟹x=-iy$$

Eigenvector: $\left[\begin{array}{c}1\\ -i\end{array}\right]$

4. Find the rotation and dilation of $A=\left[\begin{array}{cc}1& 3\\ -3& 1\end{array}\right]$.

Rotation and Dilation

$$A=\left[\begin{array}{cc}1& 3\\ -3& 1\end{array}\right]=\left[\begin{array}{cc}\sqrt{10}& 0\\ 0& \sqrt{10}\end{array}\right]\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$ $$\theta =\arctan \left(\frac{-3}{1}\right)=-\arctan (3)$$

Dilation factor: $\sqrt{10}$, Rotation angle: $-\arctan (3)$

5. Write $A=PB{P}^{-1}$ where $B$ is a rotation-dilation matrix $A=\left[\begin{array}{cc}3& 1\\ -2& 1\end{array}\right]$.

Eigenvalues and Eigenvectors

$$\det (A-\lambda I)=\left|\begin{array}{cc}3-\lambda & 1\\ -2& 1-\lambda \end{array}\right|=(3-\lambda )(1-\lambda )+2={\lambda }^2-4\lambda +5=0$$ $$\lambda =2\pm i$$

For $\lambda =2+i$:

$$\left[\begin{array}{cc}1-i& 1\\ -2& -1-i\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0⟹x=(1+i)y$$

Eigenvector: $\left[\begin{array}{c}1+i\\ 1\end{array}\right]$

For $\lambda =2-i$:

$$\left[\begin{array}{cc}1+i& 1\\ -2& -1+i\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0⟹x=(1-i)y$$

Eigenvector: $\left[\begin{array}{c}1-i\\ 1\end{array}\right]$

Matrix $P$ and $B$

$$P=\left[\begin{array}{cc}1+i& 1-i\\ 1& 1\end{array}\right],B=\left[\begin{array}{cc}2+i& 0\\ 0& 2-i\end{array}\right]$$ $$A=PB{P}^{-1}$$

6. Prove that $\lambda =a+ib$ is an eigenvalue of the matrix $A=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$.

Proof

$$\det (A-\lambda I)=\left|\begin{array}{cc}a-\lambda & -b\\ b& a-\lambda \end{array}\right|=(a-\lambda {)}^2+b^2=0$$ $$\lambda =a\pm ib$$

Thus, $\lambda =a+ib$ is an eigenvalue of $A$.